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A Givens rotation acts as the identity except on coordinates $i,j$, where its block can be writtenThe block has orthonormal columns because ${\cos }^2\theta +{\sin }^2\theta =1$ and their inner product is zero, so $J^{T}J=I$. For $r=\sqrt{v_i^2+v_j^2}>0$, choose $\cos \theta =v_i/r$, $\sin \theta =v_j/r$. The new coordinates are $(r,0)$, eliminating the $j$th entry. If both entries are zero, the identity already works.

For the specified vector, one suitable matrix isIts rows are orthonormal, so it is orthogonal; in fact its determinant is one. The positive nonzero output component must be five because an orthogonal matrix preserves Euclidean norm.

The matrix is not unique. Left multiplication by any orthogonal matrix fixing the second coordinate vector preserves this output. For example arbitrary rotations of the first and third coordinates provide infinitely many alternatives.

Let the nearest-neighbour spacing be $d$, so the primitive unit cell has length $\ell =2d$. Denote the two longitudinal displacements in cell $n$ by $u_n,v_n$, with spring $K$ within each cell and spring $K^{\prime }$ between cells. Hooke's law and Newton's second law giveFor normal modes $u_n=u{e}^{i(qn\ell -\omega t)}$ and $v_n=v{e}^{i(qn\ell -\omega t)}$, the two amplitudes satisfyIts vanishing determinant gives the alternating-spring chain dispersion relationThe minus sign is the acoustic phonon branch, with ${\omega }_{-}(0)=0$ and $u=v$ at the origin. The plus sign is the optical phonon branch, with $u=-v$ at the origin. For positive springs the acoustic phonon branch rises from zero to $\sqrt{2\min (K,K^{\prime })/m}$ at the Brillouin zone edge; the optical phonon branch falls from $\sqrt{2(K+K^{\prime })/m}$ to $\sqrt{2\max (K,K^{\prime })/m}$. Their long-wavelength limits areThus the leading frequency in the strong-bond limit of the alternating-spring chain is $\sqrt{2K/m}$, the internal stretching frequency of a pair with reduced mass $m/2$.

For one coordinate with Hamiltonian $H={\hat{p}}^2/(2m)+V(\hat{x})$, the propagator is $K(x_f,t_f;x_i,t_i)=⟨x_f|e^{-iHT/\hslash }|x_i⟩$, where $T=t_f-t_i>0$. Divide $T$ into $M$ intervals $\delta t$, use the Trotter product formula to separate kinetic and potential evolution, and insert complete position and momentum bases at each slice. The momentum integral is Gaussian:Multiplying the short-time kernels gives the time-sliced configuration-space path integral,with $x_0=x_i$, $x_M=x_f$. Continuum notation abbreviates this limit as ${\int }_{x_i}^{x_f}\mathcal{D}x{e}^{iS[x]/\hslash }$, where $S=\int dt[m{\dot{x}}^2/2-V(x)]$. The measure includes the displayed normalization; it is not an unweighted count of paths. A small damping or analytic-continuation prescription defines the oscillatory integrals. The paths contribute complex amplitudes, so alternative histories interfere before probabilities are calculated. This is the central difference from a classical statistical mixture of trajectories.^[ref]

When actions vary rapidly relative to $\hslash$, stationary phase selects classical paths. Put $x=x_{\mathrm{cl}}+\eta$, with $\eta$ zero at the endpoints. Vanishing first variation gives $m{\ddot{x}}_{\mathrm{cl}}+V^{\prime }(x_{\mathrm{cl}})=0$. The quadratic term isThe resulting Gaussian fluctuation determinant is the semiclassical propagator prefactor. In one dimension, away from caustics,The branch convention makes $\nu =0$ for a direct free-particle path before any caustic. The Maslov index tracks the phase changes as fluctuations pass through a focal singularity. In several dimensions replace the endpoint derivative by the Van Vleck determinant. The approximation sums all relevant classical paths; real-time action need only be stationary, not a minimum. It fails in this simple form at a caustic or coalescing saddles, where a uniform approximation is needed. Quadratic actions have no higher fluctuation terms, making their Gaussian evaluation exact. For example $S_{\mathrm{cl}}=m(x_f-x_i{)}^2/(2T)$ gives the exact free-particle kernel $\sqrt{m/(2\pi i\hslash T)}{e}^{i{S}_{\mathrm{cl}}/\hslash }$. Tunnelling involves classically forbidden regions and can be described by imaginary-time saddles rather than real classical trajectories.

Quantum statistical mechanics follows by Wick rotation $t=-i\tau$. The imaginary-time evolution operator is $e^{-\tau H/\hslash }$, with Euclidean actionTaking a thermal trace identifies and integrates the endpoint positions:This thermal trace as a closed Euclidean path integral explains the periodic imaginary-time direction. A density-matrix element instead retains open endpoints. For many identical particles the trace also sums over endpoint permutations, with a plus sign for bosons and permutation parity for fermions. In coherent-state language this becomes periodic bosonic fields and antiperiodic Grassmann fields, with corresponding Matsubara frequencies.

The Euclidean weight resembles a classical Boltzmann weight. A time-sliced local quantum field in $d$ spatial dimensions becomes a classical-looking field model in $d+1$ dimensions, with anisotropic couplings and extra extent $\beta \hslash$. Thus zero-temperature quantum criticality can be treated using an infinitely extended imaginary-time direction, while nonzero temperature imposes a finite thickness. The dynamical scaling of time need not equal the spatial scaling: the dynamical critical exponent specifies their relationship. This quantum-to-classical imaginary-time mapping requires qualification when weights have signs or imaginary terms; such a functional integral is not automatically an ordinary positive classical probability distribution. Even with a positive weight, the mapping concerns equilibrium statistics, not literal dissipative real-time trajectories. For a free particle the Euclidean kernel is a diffusion kernel with diffusion coefficient $\hslash /(2m)$, illustrating the mathematical correspondence.

The classical high-temperature limit is a further approximation, rather than simply another name for imaginary time. If quantum level spacings and exchange effects are negligible, a one-particle partition function in $d$ dimensions becomesFor indistinguishable particles there is a factor $1/N!$ in the dilute classical gas, and exchange is negligible when the number density times the thermal de Broglie wavelength to power $d$ is small. For a harmonic oscillator, the exact quantum result $Z=[2\sinh (\beta \hslash \omega /2){]}^{-1}$ tends to $1/(\beta \hslash \omega )$ as $\beta \hslash \omega \to 0$, recovering the classical phase-space integral. stationary phase explains classical mechanics; the short imaginary-time, weak-exchange limit explains classical statistical mechanics. These are related but distinct limits of the same quantum formulation.

At wavelengths with $q^2/(4m{\rho }_0)\ll g$, neglect the density-gradient term in the quadratic density-phase action. Complete the square in its remaining density terms:The convergent Gaussian integral can be performed by a contour shift or by its Fourier-transform identity. Its determinant is independent of $\phi$ at this order. Combining that normalization with the saddle action into a constant $S_0$ givesThe saddle contribution to $S_0$ is $-\beta V{\mu }^2/(2g)$; the additional measure-dependent constant need not be assigned an invented numerical value. The density Gaussian extends the fluctuation integration beyond its exact positivity domain, which is justified only while fluctuations around ${\rho }_0>0$ are small. The smooth zero-winding assumption is also required as discussed above. A constant global phase zero mode gives a finite compact-group volume, handled separately from nonzero phonon modes.

Fourier transforming the Bose-gas phase-only action gives a kernel ${\omega }_n^2/g+({\rho }_0/m)q^2$. Analytically continuing ${\omega }_n\mapsto -i(\omega +i{0}^{+})$ gives its real-frequency poles,In the present units the excitation energy equals $\omega$. Each real Fourier coordinate is a harmonic oscillator with inertia coefficient $1/g$ and spring coefficient ${\rho }_0{q}^2/m$, hence frequency equal to the square root of their ratio. This is the direct relation to a quantum harmonic chain. The mode is a gapless sound phonon; its finite speed reflects both the compressibility $\partial \rho /\partial \mu =1/g$ and the phase stiffness ${\rho }_0/m$.

Retaining the density-gradient term provides a useful check. Its integration instead gives a temporal coefficient $1/A_q$, soThis is the Bogoliubov spectrum. Its small-$q$ limit is the derived sound branch; at large $q$ within the contact model it crosses towards particle-like behaviour. The crossover in the squared dispersion is $q\sim 2\sqrt{mg{\rho }_0}$, precisely the scale beyond which the discarded density gradient is not small.

The same action checks the dimensional restriction in the saddle analysis. At nonzero temperature its zero Matsubara mode gives phase fluctuations proportional to $\int d^{d}q/q^2$; at zero temperature integrating the frequency gives $\int d^{d}q/|q|$. Their infrared divergences invalidate uniform true long-range order for $d\le 2$ at nonzero temperature and for $d=1$ at zero temperature. The long-wave collective description can remain useful for correlations even where a global condensate saddle is not a literal symmetry-broken state. The contact coupling here is an effective low-energy parameter; this calculation does not determine ultraviolet renormalization or trap-specific surface modes.

Write $u_k=\cosh {\theta }_k$, $v_k=\sinh {\theta }_k$, taking real even ${\theta }_k$. The transformation givesThe second equality uses $u_{-k}=u_k$ and $v_{-k}=v_k$, while the first uses $u_k^2-v_k^2=1$. The creation-creation commutator vanishes by conjugation. Therefore the $b$ modes are canonical bosons. In the paired basis,Its determinant is one and its inverse reverses the signs of the off-diagonal entries. Hence $\boxed{a_k=u_k{b}_k-v_k{b}_{-k}^{†}}$. At self-inverse momenta $k=0,\pi$ this is the corresponding one-mode squeezing transformation, still canonical for finite ${\theta }_k$.

For one bosonic mode, define the unnormalized bosonic coherent state $|z⟩=e^{z{a}^{†}}|0⟩$. The commutator $[a,a^{†}]=1$ gives $a|z⟩=z|z⟩$ and $⟨z^{\prime }|z⟩=e^{{\bar{z}}^{\prime }z}$. The normalized state is $e^{-|z{|}^2/2}|z⟩$; it is not orthogonal to a different coherent state. The coherent-state resolution of identity follows from Gaussian moments,For many modes take products of these states and measures. The complex amplitudes have the natural scale of square root occupation; coherent states turn normal-ordered matrix elements into classical symbols without replacing the quantum trace by a single classical configuration.

For fermions, $\{c_i,c_j^{†}\}={\delta }_{ij}$ and a nilpotent annihilation operator cannot have an arbitrary nonzero ordinary complex eigenvalue. Introduce independent Grassmann variables ${\eta }_i,{\bar{\eta }}_i$, anticommuting among themselves and with the fermionic operators, and define $|\eta ⟩=\exp (-{\sum }_i{\eta }_i{c}_i^{†})|0⟩$. Then $c_i|\eta ⟩={\eta }_i|\eta ⟩$ and $⟨{\bar{\eta }}^{\prime }|\eta ⟩=e^{{\sum }_i{\bar{\eta }}_i^{\prime }{\eta }_i}$. With an ordered Berezin integral measure satisfying $\int d\bar{\eta }d\eta \bar{\eta }\eta =-1$ for one pair,for an even operator $O$; this is the fermionic coherent-state trace. In one mode the trace integrand is $e^{-\bar{\eta }\eta }(O_{00}-\bar{\eta }\eta O_{11})$, whose integral is $O_{00}+O_{11}$. Without the sign reversal it would be the supertrace, not the thermal trace. These formal Grassmann-valued coherent states do not represent ordinary complex-valued fermion amplitudes.

To construct the many-body thermal path integral, write $\mathcal{Z}=\mathrm{Tr}{e}^{-\beta K}$ with $K=H-\mu \hat{N}$ and divide the imaginary-time interval $\mathcal{T}=\beta \hslash$ into slices of width $\delta \tau$. Insert the appropriate coherent-state resolution of identity between factors. For a normal-ordered Hamiltonian, the adjacent-slice kernel iscoherent-state time slicing makes the overlap and Gaussian measure combine into the first-order time derivative. The grand-canonical coherent-state action in the regulated continuum limit isFor bosons the integration variables are complex, with the conjugate labels on the original integration contour; for fermions the barred and unbarred Grassmann variables are independent. The continuum notation retains the time-sliced ordering prescription. For example, $n(n-1)=a^{†2}{a}^2$ has normal symbol ${\bar{z}}^2{z}^2$, whereas $n^2$ has that symbol plus $\bar{z}z$. Dropping that distinction changes a theory. The coherent-state thermal boundary conditions follow because the bosonic trace closes paths periodically and the fermionic sign in the trace closes them antiperiodically:The barred fields obey the same respective boundary conditions. Hence Matsubara frequencies are $2\pi n/(\beta \hslash )$ and $(2n+1)\pi /(\beta \hslash )$. The derivative term encodes the canonical structure; omitting it would remove the quantum dynamics.^[ref]

As a check, a free diagonal mode with $K=\xi a^{†}a$ gives $Z_B=(1-e^{-\beta \xi }{)}^{-1}$ for $\xi >0$, whereas $K=\xi c^{†}c$ gives $Z_F=1+e^{-\beta \xi }$. Gaussian integration produces an inverse determinant for bosons and a determinant for independent complex fermions, with their respective temporal boundary conditions. Differentiating $\log Z$ recovers Bose-Einstein and Fermi-Dirac occupations. Interactions become higher powers in the normal-ordered symbols, and sources generate time-ordered correlations. The coherent-state construction preserves statistics, ordering and boundary conditions together; a formal continuum expression without its measure and temporal prescription is incomplete.

Use the contact-interaction Bose-gas field integral for a homogeneous three-dimensional weakly interacting Bose gas with repulsive short-range interaction as the standard model. Let particle mass be $m_b$, scattering length $a_s>0$, physical coupling $g=4\pi {\hslash }^2{a}_s/m_b$ and gas parameter $n{a}_s^3\ll 1$. These are symbolic model definitions, not supplied numerical parameters. The grand-canonical coherent-state integral has periodic fields and actionThe contact term is the normal symbol of $g{\hat{\psi }}^{†2}{\hat{\psi }}^2/2$. In three dimensions a contact interaction needs a regulator and coupling matched to the measured scattering length; the displayed $g$ is the low-energy physical coupling. This field integral is useful because an ordered saddle and its quantum/thermal fluctuations follow from the same action.

Stationarity with respect to $\bar{\psi }$ gives the imaginary-time Gross–Pitaevskii equation. A uniform saddle exists at $\mu >0$ with ${\psi }_0=\sqrt{n_0}{e}^{i{\theta }_0}$ and condensate density $n_0=\mu /g$. Its arbitrary phase reflects global $U(1)$ symmetry; a selected condensate fixes a phase in the thermodynamic symmetry-breaking description. Write $\psi =\sqrt{n_0}+\eta$ after choosing that phase. Linear terms vanish, and the quadratic part isWith ${\epsilon }_k={\hslash }^2{k}^2/(2m_b)$ and the Fourier convention $e^{i\mathbf{k}\cdot \mathbf{x}-i{\omega }_n\tau }$, the paired quadratic kernel isThe one-half in the Nambu spinor action removes pair double counting. Analytic continuation of the inverse kernel gives the Bogoliubov dispersionAt high momentum it approaches the free-particle kinetic energy plus its mean-field shift. The crossover scale is the healing length $\xi =\hslash /\sqrt{2m_{b}g{n}_0}$. Gaussian integration adds a fluctuation determinant to the saddle free energy. Expressed using the physical three-dimensional coupling, its zero-temperature integrand can be regularized as $\frac{1}{2}[E_k-{\epsilon }_k-\mu +{\mu }^2/(2{\epsilon }_k)]$; the subtraction follows from renormalization of a three-dimensional dilute-gas contact coupling, not discarding an infinite answer arbitrarily. The thermal contribution is $k_{B}T\int d^{3}k/(2\pi {)}^3\log (1-e^{-\beta E_k})$. A calculation beyond mean field must keep that matching and its thermodynamic derivatives consistent.

The density-phase action of a Bose gas makes the physical content clear, $\psi =\sqrt{n_0+\delta n}{e}^{i\theta }$. Up to a periodic density total derivative, the long-wavelength quadratic action containsnumber-phase conjugacy pairs the density fluctuation and phase in the quadratic density-phase action. Dropping the density-gradient term only in the long-wave limit and completing the density Gaussian square gives the Bose-gas phase-only actionThis has the same sound pole. A term $i\hslash n_0{\partial }_{\tau }\theta$ is also present before restricting to the smooth zero-winding sector; although locally a total derivative, it cannot in general be ignored in topologically nontrivial sectors. The density and phase variables describe the same coupled Bogoliubov branch, not a second independent massive particle simply because the density has a static restoring cost.^[ref]

A phase gradient carries superflow ${\mathbf{v}}_s=(\hslash /m_b)\nabla \theta$. Its energy cost defines the superfluid number density. At leading mean-field order the stiffness uses $n_0$, but condensate density and superfluid number density are different observables: for a Galilean-invariant continuum gas at zero temperature the latter is the full density even when interactions deplete the condensate. Bogoliubov coefficients give $v_k^2=[({\epsilon }_k+g{n}_0)/E_k-1]/2$ and a small zero-temperature depletion of relative order $\sqrt{n{a}_s^3}$ in three dimensions. Large depletion or strong long-wavelength phase fluctuations would invalidate the small-fluctuation saddle approximation.

The Landau criterion for creating excitations is $v_L={\inf }_{k>0}{E}_k/(\hslash k)=c_s$ for this homogeneous dispersion. Thus repulsion produces a linear sound mode and a positive critical velocity from the Landau criterion, whereas an ideal quadratic gas has zero value of this criterion. Vortices, obstacles and nucleation can reduce an actual flow threshold below $c_s$; the criterion alone is not a complete theory of metastable flow. At finite temperature thermal quasiparticles reduce the superfluid number density. Dimensionality and finite size also matter: the uniform three-dimensional weak-depletion condensate expansion must not be transferred unchanged to a one- or two-dimensional thermodynamic system. The field integral provides systematic corrections only within its dilute, stable and sufficiently weak-fluctuation regime.

Continue the free-particle propagator to $t=-i\mathcal{T}$ and replace its coordinate mass by the angular inertia $I$. The Euclidean free-particle bridge amplitude at coincident endpoints isThis normalization includes the time-sliced Gaussian measure, not just a formal determinant with an unspecified multiplicative constant. Set $A=2{\pi }^{2}I/({\hslash }^2\beta )>0$. The correct Poisson summation formula is ${\sum }_{m}F(m)={\sum }_n{\int }_{-\infty }^{\infty }dxF(x)e^{-2\pi inx}$; the source PDF omits the integral sign before its $d\theta$. For $F(x)=e^{-A{x}^2+i2\pi fx}$ the Gaussian Fourier integral givesThe overall prefactor cancels exactly: $2\pi Z_0\sqrt{\pi /A}=1$, and ${\pi }^2/A=\beta {\hslash }^2/(2I)$. Hencein agreement with the angular-momentum trace. Pairing $m$ and $-m$ makes the winding expression real, and the momentum expression shows positivity. Both are periodic in flux and invariant under $f\mapsto -f$. At high temperature the zero-winding term gives $Z\simeq \sqrt{2\pi I/(\beta {\hslash }^2)}$, with exponentially small flux dependence; at low temperature the nearest integer angular-momentum state dominates, with two equal lowest terms at half-integer flux. These independent limits also check the normalization and the flux phase.

Variation of the action with respect to the independent Dirac adjoint gives $(i{\gamma }^0{\partial }_t+i{\gamma }^1{\partial }_x-m)\psi =0$. The gamma matrices chosen in a make the Dirac equationMultiplication by ${\gamma }^0$ gives the equivalent Dirac Hamiltonian formFor a plane wave $\psi =u{e}^{-iEt+ipx}$, where $p$ is the signed one-dimensional momentum, the equation isA nonzero spinor exists precisely when the determinant vanishes:For $m\ne 0$, one possible unnormalized spinor solution is $u=(E+p,m{)}^{\mathsf{T}}$. Alternatively, $(p{\sigma }_3+m{\sigma }_1{)}^2=(p^2+m^2)1$ follows from the Pauli matrix multiplication law, proving the same relativistic energy-momentum relation. Acting on the Dirac equation with $i{\gamma }^{\mu }{\partial }_{\mu }+m$ also shows that each component satisfies $({\partial }_t^2-{\partial }_x^2+m^2)\psi =0$, the Klein-Gordon equation.

A continuous Lorentz transformation in one spatial dimension must preserve $t^2-x^2$. In the component connected to the identity, take the image of the time axis to have velocity $v$, with $|v|<1$, and write $B_v(1,0{)}^{\mathsf{T}}=\gamma (1,v{)}^{\mathsf{T}}$. Preservation of the Minkowski norm gives ${\gamma }^2(1-v^2)=1$, so $\gamma =(1-v^2{)}^{-1/2}$. The spacelike column must be orthogonal to the timelike column and have squared Minkowski norm $-1$; continuity fixes it to $\gamma (v,1{)}^{\mathsf{T}}$. ThusHere $x=(t,x{)}^{\mathsf{T}}$ denotes the coordinate pair. This is the positive off-diagonal convention in the source; describing a passive frame moving at $+v$ instead reverses the sign of $v$.

Define the rapidity $\alpha =\mathrm{artanh}v$. The hyperbolic function identities giveIn light-cone coordinates, $t^{\prime }+x^{\prime }=e^{\alpha }(t+x)$ and $t^{\prime }-x^{\prime }=e^{-\alpha }(t-x)$. Therefore the derivatives transform oppositely:Use the field-transformation convention ${\psi }^{\prime }(x^{\prime })=S(\alpha )\psi (x)$. The component action in c is invariant underIndeed, the right-component kinetic term gains $e^{\alpha /2}{e}^{-\alpha }{e}^{\alpha /2}=1$, and the left-component kinetic term gains $e^{-\alpha /2}{e}^{\alpha }{e}^{-\alpha /2}=1$. The two factors in each Dirac mass term cancel as well. Finally $\det B=1$, so $d{t}^{\prime }d{x}^{\prime }=dtdx$ and the whole action is unchanged. These are the chiral boost weights in two spacetime dimensions.

In spinor notation, $S(\alpha )=\exp (\alpha {\gamma }^5/2)$ satisfies $S^{-1}{\gamma }^{\mu }S=B^{\mu }{}_{\nu }{\gamma }^{\nu }$ and $S^{†}{\gamma }^{0}S={\gamma }^0$, so ${\bar{\psi }}^{\prime }=\bar{\psi }{S}^{-1}$. These identities directly verify covariance of the Dirac equation and the Lorentz invariance of its action. The matrices $S$ are not unitary with respect to the ordinary component norm; the Dirac adjoint supplies the invariant bilinear.

The statement about there being only Lorentz boosts applies to the identity component of the One-dimensional Lorentz group. The full Lorentz group also has discrete transformations: the concrete spatial reflection $\mathrm{diag}(1,-1)$ preserves the Minkowski metric but has determinant $-1$, so it cannot equal any displayed $B(\alpha )$, whose determinant is $+1$.

A density operator is a positive self-adjoint trace-class operator with $\boxed{\mathrm{Tr}\rho =1}$. A finite expectation is $\boxed{⟨A{⟩}_{\rho }=\mathrm{Tr}(\rho A)}$. It represents a pure state exactly when $\rho =|v⟩⟨v|$ for a unit vector, equivalently ${\rho }^2=\rho$ or $\mathrm{Tr}{\rho }^2=1$.

For a finite-dimensional spectral decomposition $\rho ={\sum }_j{p}_j|v_j⟩⟨v_j|$, introduce an auxiliary space with orthonormal vectors $f_j$ and use the enlarged space $\mathcal{K}=\mathcal{H}\otimes \mathcal{E}$. DefineThen $\Vert {\Omega }_{\rho }{\Vert }^2=\sum p_j=1$ and $⟨{\Omega }_{\rho },\tilde{A}{\Omega }_{\rho }⟩=\sum p_j⟨v_j,A{v}_j⟩=\mathrm{Tr}(\rho A)$. This purification shows that subsystem mixedness can arise from entanglement with an unobserved environment even when the total state is pure. Subsystem measurements alone cannot distinguish these two descriptions.

A nonzero pure bipartite vector is entangled when it cannot be written as a tensor product of two vectors; equivalently its Schmidt rank exceeds one. For mixed bipartite states, entanglement means absence of a convex decomposition into product density operators.

For the nonorthogonal expansion in the question, the unnormalized reduced operator isIndeed $⟨\Psi ,(A\otimes I)\Psi ⟩={\sum }_{j,k}⟨{\xi }_k,A{\xi }_j⟩⟨{\eta }_k,{\eta }_j⟩=\mathrm{Tr}(A{S}_1)$. Positivity follows because $⟨v,S_{1}v⟩=\Vert {\sum }_j⟨v,{\xi }_j⟩{\eta }_j{\Vert }^2\ge 0$, and $\mathrm{Tr}{S}_1=\Vert \Psi {\Vert }^2=:Z$. Thus $\boxed{{\rho }_1=S_1/Z}$ is a density operator; if $\Psi$ is already normalized, $Z=1$. The Gram factors cannot be dropped merely because the vectors are called bases. For unnormalized $\Psi$, physical expectations likewise divide its raw quadratic form by $Z$.

For the Schmidt decomposition diagonalize $S_1$ in an orthonormal basis ${\xi }_j$ with its positive eigenvalues $c_j^2$. Expand $\Psi ={\sum }_j{\xi }_j\otimes w_j$ with $w_j=(⟨{\xi }_j|\otimes I)\Psi$. The reduced matrix gives $⟨w_k,w_j⟩=c_j^2{\delta }_{jk}$. Therefore ${\eta }_j=w_j/c_j$ are orthonormal andZero-eigenvalue components vanish because their squared norms are zero. This proves the decomposition for an arbitrary nonzero vector; the zero vector uses the empty sum. For a unit vector $\sum c_j^2=1$.

In the numerical example the coefficient matrix in the two product bases is $C=\left(\begin{array}{ccc}1& 2& 3\\ 0& 1& 2\end{array}\right)$. Taking the first-subsystem trace yieldsThe vector $(1,-2,1{)}^{\mathsf{T}}$ is in its nullspace, so rank is at most two; its leading $2\times 2$ minor has determinant one, so $\boxed{\mathrm{rank}{S}_2=2}$. This is the rank of $C$ and the Schmidt rank, proving entanglement. The displayed source ket has norm squared $19$, and the displayed reduced matrix has trace $19$. The physical density operator is $\boxed{{\rho }_2=S_2/19}$; its nonzero eigenvalues are $(19\pm \sqrt{337})/38$, since $C{C}^{\mathsf{T}}=\left(\begin{array}{cc}14& 8\\ 8& 5\end{array}\right)$. The raw matrix is correct for the unnormalized quadratic form actually printed.

A density operator is a positive self-adjoint trace-class operator with $\boxed{\mathrm{Tr}\rho =1}$. A finite expectation is $\boxed{⟨A{⟩}_{\rho }=\mathrm{Tr}(\rho A)}$. It represents a pure state exactly when $\rho =|v⟩⟨v|$ for a unit vector, equivalently ${\rho }^2=\rho$ or $\mathrm{Tr}{\rho }^2=1$.

For a finite-dimensional spectral decomposition $\rho ={\sum }_j{p}_j|v_j⟩⟨v_j|$, introduce an auxiliary space with orthonormal vectors $f_j$ and use the enlarged space $\mathcal{K}=\mathcal{H}\otimes \mathcal{E}$. DefineThen $\Vert {\Omega }_{\rho }{\Vert }^2=\sum p_j=1$ and $⟨{\Omega }_{\rho },\tilde{A}{\Omega }_{\rho }⟩=\sum p_j⟨v_j,A{v}_j⟩=\mathrm{Tr}(\rho A)$. This purification shows that subsystem mixedness can arise from entanglement with an unobserved environment even when the total state is pure. Subsystem measurements alone cannot distinguish these two descriptions.

A nonzero pure bipartite vector is entangled when it cannot be written as a tensor product of two vectors; equivalently its Schmidt rank exceeds one. For mixed bipartite states, entanglement means absence of a convex decomposition into product density operators.

For the nonorthogonal expansion in the question, the unnormalized reduced operator isIndeed $⟨\Psi ,(A\otimes I)\Psi ⟩={\sum }_{j,k}⟨{\xi }_k,A{\xi }_j⟩⟨{\eta }_k,{\eta }_j⟩=\mathrm{Tr}(A{S}_1)$. Positivity follows because $⟨v,S_{1}v⟩=\Vert {\sum }_j⟨v,{\xi }_j⟩{\eta }_j{\Vert }^2\ge 0$, and $\mathrm{Tr}{S}_1=\Vert \Psi {\Vert }^2=:Z$. Thus $\boxed{{\rho }_1=S_1/Z}$ is a density operator; if $\Psi$ is already normalized, $Z=1$. The Gram factors cannot be dropped merely because the vectors are called bases. For unnormalized $\Psi$, physical expectations likewise divide its raw quadratic form by $Z$.

For the Schmidt decomposition diagonalize $S_1$ in an orthonormal basis ${\xi }_j$ with its positive eigenvalues $c_j^2$. Expand $\Psi ={\sum }_j{\xi }_j\otimes w_j$ with $w_j=(⟨{\xi }_j|\otimes I)\Psi$. The reduced matrix gives $⟨w_k,w_j⟩=c_j^2{\delta }_{jk}$. Therefore ${\eta }_j=w_j/c_j$ are orthonormal andZero-eigenvalue components vanish because their squared norms are zero. This proves the decomposition for an arbitrary nonzero vector; the zero vector uses the empty sum. For a unit vector $\sum c_j^2=1$.

In the numerical example the coefficient matrix in the two product bases is $C=\left(\begin{array}{ccc}1& 2& 3\\ 0& 1& 2\end{array}\right)$. Taking the first-subsystem trace yieldsThe vector $(1,-2,1{)}^{\mathsf{T}}$ is in its nullspace, so rank is at most two; its leading $2\times 2$ minor has determinant one, so $\boxed{\mathrm{rank}{S}_2=2}$. This is the rank of $C$ and the Schmidt rank, proving entanglement. The displayed source ket has norm squared $19$, and the displayed reduced matrix has trace $19$. The physical density operator is $\boxed{{\rho }_2=S_2/19}$; its nonzero eigenvalues are $(19\pm \sqrt{337})/38$, since $C{C}^{\mathsf{T}}=\left(\begin{array}{cc}14& 8\\ 8& 5\end{array}\right)$. The raw matrix is correct for the unnormalized quadratic form actually printed.